Design of bracing

1. CB523
BRACING DESIGN
ENG. ESRAA HUSSEIN MOHAMED

2. BRACING:
• IS A STRUCTURAL MEMBER USED TO SUPPORT ANOTHER.
• IT USED TO RESIST THE MOVEMENT OF FORMWORK AND WIND LOADS.
Bracer

3. STEPS OF SOLUTIONS:-
• L= ℎ′! + 𝑙′!
L
l’
h’
h
Where:-
h : height of form (ft) [m]
h’: height of top of strut (ft) [m]
L : length of strut (ft) [m]
l’: horizontal distance form the bottom of strut (ft) [m]

4. Strut Load per foot of form
• P’=
"∗!∗$
%!∗!&
Where:-
P : strut load per foot of form (lb/ft) [kN/m]
H : lateral load at top of form (lb/ft) [kN/m]
h : height of form (ft) [m]
h’: height of top of strut (ft) [m]
L : length of strut (ft) [m]
l’: horizontal distance from form to bottom of strut (ft) [m]

5. •
!
"
< 50
•
!
"
> 50
Safe
Un Safe
Strut
Lnew=
"
#
L/2
L/2
Check of Buckling:-
D
L

6. Allowable Compressive Stress:
• FC
’=".$ %
(
!
"
)#
“PSI” <FC <FT Safe

7. Maximum Allowable Compressive Force:
• P = B*D*FC
’
Maximum Strut Spacing:
• S = ⁄(
()
“Lb”
“ft”

8. EXAMPLE:-
v Determine the maximum spacing of 2”x4” lateral braces for the wall form of 8 ft
high. Lateral wood bracing will be attached at a height of 6ft above the form
bottom and anchored 5ft away from the bottom of the form. Design wind load
is 25 psf.
- Wood charactesrstics: Fc = 850 PSI, Ft = 725 PSI,
E =1,400,000 PSI
h= 8ft
h’= 6ft
l’= 5ft
Solution:-
L= ℎ′! + 𝑙′! = 6! + 5! = 7.81ft
P’=
" ∗ $ ∗ %
&! ∗ $'
=
()) ∗ * ∗ +.*(
- ∗ .
= 208.27 lb/ft
L
5ft
6ft
8ft

9. h= 8ft àFrom (Table 13-3) H=100lb/ft at least
$∗/0
!
=
*∗!-
!
= 100lb/ft
P’=
" ∗ $ ∗ %
&! ∗ $'
=
()) ∗ * ∗ +.*(
- ∗ .
= 208.27 lb/ft
%
1
=
+.*( ∗(!
(.-
= 62.48 > 50
Lnew=
+.*( ∗(!
!
= 46.86 in
%!"#
1
=
2..*.
(.-
= 31.24 < 50
Un Safe
Safe

10. - Wood charactesrstics: Fc= 850 PSI, Ft = 725 PSI, E =1,400,000 PSI
• Fc
’=
).3 ∗(,2)),)))
(
"#.%#
&.'
)(
= 430.355 PSI
<Fc= 850PSI
<Ft=725PSI
• P = 3.5 * 1.5 * 430.355= 2259.264 Lb
• S =
7
7'
=
!!-8.!.2
!)*.!+
= 10.85 ft
“ Safe ”

11. THANK YOU